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If $$y^4+5xy+y =2$$ find $\frac{dy}{dx}$ and d^2y/dx^2 when $x=0$ and $y=1$

I could not differentiate w.r.t x...answer coming to 0 which is not the correct answer.

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$y^4+5xy+y=2$

$\implies \frac d{dx} (y^4+5xy+y)=0$

$\implies 4y^3 \frac {dy}{dx}+5(y+x\frac {dy}{dx})+\frac {dy}{dx}=0 $

$\implies (4y^3+5x+1)\frac {dy}{dx}+5y=0$

Putting the values of $x$ & $y$

$\implies 5\frac {dy}{dx}+5=0$

$\implies \frac {dy}{dx}=-1$

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We have $4y^3 \frac{dy}{dx}+5y+5x\frac{dy}{dx}+\frac{dy}{dx}=0$ and plugging $x=0$ and $y=1$ we get $4\frac{dy}{dx}+5+\frac{dy}{dx}=0$ and thus $\frac{dy}{dx}=-1$

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Alternatively, refer to the inverse function: $$y^4+5xy+y =2 \iff x=\frac{2-y-y^4}{5y}=\frac2{5y}-\frac15-\frac{y^3}{5}$$ Hence: $$\frac{dy}{dx}|_{x=0}=\frac1{\frac{dx}{dy}|_{y=1}}=\frac1{\left(-\frac25-\frac{3y^2}{5}\right)|_{y=1}}=\frac1{-1}=-1.$$

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